University Quantum Field Theory I · Fermions in Perturbation Theory: the Dirac Propagator, Fermion Signs and Yukawa Theory · 7.9

Pseudoscalar exchange: a spin-spin and a tensor force

Scalar exchange attracted every pair and never asked about spin. Put a γ⁵ at each vertex and the same diagram reads the spins instead: σ⋅q at each end, two gradients of a Yukawa function, and a tensor force that pushes a polarised pair apart end to end and pulls it together side by side. This is the skeleton of one-pion exchange.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Pseudoscalar exchange is Yukawa theory with iγ⁵ at each vertex, and that one matrix changes what the force acts on. γ⁵ is off-diagonal between the large and small components of a Dirac spinor, so the slow-fermion bilinear ū(p′)γ⁵u(p) is not the scalar's 2m but χ′†σ⋅(p − p′)χ: a spin operator of size |q|, the only pseudoscalar linear in the momentum transfer that a slow spin-½ particle can build, which is why the boson leaves the vertex in a P wave. The tree amplitude for ψψ → ψψ therefore carries (σ₁⋅q)(σ₂⋅q) where scalar exchange carried 4m², and Born matching gives Ṽ(q) = −(g²/4m²)(σ₁⋅q)(σ₂⋅q)/(|q|² + m_φ²). Each q is a gradient in position space, so V(r) is two gradients of a Yukawa function, and separating their trace from their traceless part yields a contact term, a spin-spin Yukawa term in σ₁⋅σ₂, and a tensor term in S₁₂ whose radial factor 1 + 3/x + 3/x² makes it the strongest piece wherever the force is strong.

The force averages to zero over unpolarised spins, attracts or repels a polarised pair according to its orientation, and becomes a dipole-dipole law as the boson's mass goes to zero. It is the skeleton of one-pion exchange and the source of the nuclear tensor force, and it stops at its own assumptions: it is the leading term in |q|/m, its fermions are points, and its 1/r³ core and δ-function sit exactly where heavier exchanges and the nucleons' size take over.

Simple definition: Pseudoscalar exchange is the force between two fermions that trade a spin-0 boson coupled through iψ̄γ⁵ψ; for slow fermions each vertex becomes σ⋅q, so the force acts on their spins, as a spin-spin term plus a tensor term in S₁₂.
Example: Two fermions of mass 939 MeV, 1.5 fm apart with both spins along z, exchanging a 135 MeV pseudoscalar with g²/4π = 13.6: V = +16.1 MeV end to end, −6.4 MeV side by side, and −3.3 MeV in the spin singlet.

Pseudoscalar density of a slow fermionū(p′)γ⁵u(p) = χ′†σ⋅(p − p′)χ

No χ′†χ term survives: against the scalar's 2m it is a spin operator of relative size |q|/2m, built from one spinor's small components.

Exact for E′ = E. p, p′ three-momenta in GeV; χ, χ′ two-component spin states; the bilinear in GeV, like ūu = 2m

Momentum-space potential, derivedṼ(q) = −(g²/4m²)(σ₁⋅q)(σ₂⋅q)/(|q|² + m_φ²)

Its trace over both spins is zero, so an unpolarised pair feels no force on average, where scalar exchange attracts every pair.

g dimensionless; m, m_φ and q in GeV; Ṽ in GeV⁻², an energy times a volume (ħ³c³ turns it into MeV fm³)

Spin-spin and tensor Yukawa termsV(r) = (g²/4π)(m_φ²/12m²)[σ₁⋅σ₂ + S₁₂T(x)]e^(−x)/r, T(x) = 1 + 3/x + 3/x²

T exceeds 1 at every r and grows as 3/x², so the tensor term outweighs the spin-spin term wherever the force is strong.

x = m_φr, dimensionless; V in GeV. For laboratory numbers x = m_φc²r/ħc with ħc = 197.327 MeV fm

Contact termV_δ(r) = −(g²/12m²) σ₁⋅σ₂ δ³(r)

The constant that the trace part of Ṽ(q), −(g²/12m²)σ₁⋅σ₂|q|²/(|q|² + m_φ²), tends to at large |q|. It acts only where the two fermions overlap, which a model of point fermions cannot describe.

δ³(r) in GeV³ and g²/m² in GeV⁻², so V_δ is an energy; it comes from ∇²Y = m_φ²Y − δ³(r), Y = e^(−m_φr)/4πr

Tensor operatorS₁₂ = 3(σ₁⋅r̂)(σ₂⋅r̂) − σ₁⋅σ₂

It averages to zero over directions of r̂, so it does nothing in a pure S wave at first order: it works by coupling ³S₁ to ³D₁.

Dimensionless. Zero on the singlet; ⟨S₁₂⟩ = 3cos²θ − 1 for both spins along z with r̂ at angle θ to z

Massless limit: a dipole-dipole lawV → (g²/4π)S₁₂/(4m²r³) as m_φr → 0

A spin-0 field attracts like sources, so these dipoles repel end to end, where two bar magnets attract.

Plus the contact term. Two Dirac magnetic moments give −(α/4m²)S₁₂/r³: the same shape, with −α in place of g²/4π

γ⁵ joins large components to small ones

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Take Lᵢₙₜ = −igφψ̄γ⁵ψ: ψ̄γ⁵ψ is anti-Hermitian, so the i makes the coupling Hermitian, and the vertex is gγ⁵ where the scalar's was −ig; only g² will matter. In the Dirac representation (UAQM-04, quoted) u(p) = (√(E + m) χ, σ⋅p χ/√(E + m)), and γ⁰γ⁵ is block off-diagonal, so ū(p′)γ⁵u(p) always multiplies the large components of one spinor by the small components of the other: χ′†[√((E′ + m)/(E + m)) σ⋅p − √((E + m)/(E′ + m)) σ⋅p′]χ, derived. In elastic centre-of-mass scattering E′ = E, and it is exactly χ′†σ⋅(p − p′)χ. No χ′†χ term survives. The scalar bilinear weighs a slow fermion as 2m; this one is a spin operator of size |q|, a relative 0.11 for |q| = 0.21 GeV on a 0.938 GeV nucleon. Parity demands it: a pseudoscalar field must couple to a pseudoscalar, and the only one linear in q a slow fermion can offer is σ⋅q, σ being an axial vector and q a polar one. The boson, of intrinsic parity −1, leaves the vertex in a P wave.

Two sign changes turn 4m² into (σ₁⋅q)(σ₂⋅q)

Order the operators as topic 7.05 did, so the direct contraction for ψψ → ψψ carries +1 (quoted) and iMₜ = (ū₃gγ⁵u₁)(ū₄gγ⁵u₂) i/(t − m_φ²). The u-channel diagram, with its relative −1, is the exchange term of the same potential between identical fermions, so V is read from the t-channel alone. Against scalar exchange two signs change. The vertices give (gγ⁵)², a +g² where the scalar's (−ig)² gave −g². And the momentum runs opposite ways through the two vertices: with q = p₃ − p₁ = p₂ − p₄, vertex 1 gives −χ₃†σ⋅qχ₁ and vertex 2 gives +χ₄†σ⋅qχ₂. In elastic centre-of-mass kinematics q⁰ = 0 and t = −|q|², so Mₜ = g²(σ₁⋅q)(σ₂⋅q)/(|q|² + m_φ²), derived: the scalar's 4m²g²/(|q|² + m_φ²) with 4m² replaced by a spin operator. Born matching, Ṽ = −Mₜ/4m² (topic 7.05, quoted), gives Ṽ(q) = −(g²/4m²)(σ₁⋅q)(σ₂⋅q)/(|q|² + m_φ²), derived. Since every Pauli matrix is traceless, the trace over both spins vanishes: averaged over spins, the pair feels no force at all.

Fourier transform: two gradients of a Yukawa function

Each qᵢ acts as −i∂ᵢ on e^(iq⋅r), so (σ₁⋅q)(σ₂⋅q) becomes −(σ₁⋅∇)(σ₂⋅∇) and V(r) = (g²/4m²)(σ₁⋅∇)(σ₂⋅∇)Y(r), with Y = e^(−m_φr)/4πr the transform of 1/(|q|² + m_φ²), quoted. Split ∂ᵢ∂ⱼY into a traceless part and a trace: (r̂ᵢr̂ⱼ − δᵢⱼ/3)(Y″ − Y′/r) + (δᵢⱼ/3)∇²Y. For this Y, Y″ − Y′/r = m_φ²Y(1 + 3/x + 3/x²) with x = m_φr, and ∇²Y = m_φ²Y − δ³(r), the δ-function coming from the Coulomb-like 1/4πr singularity at the origin. Contracting with σ₁ᵢσ₂ⱼ turns the traceless tensor into S₁₂/3 and δᵢⱼ/3 into σ₁⋅σ₂/3, so V(r) = (g²/4π)(m_φ²/12m²)[σ₁⋅σ₂ + S₁₂(1 + 3/x + 3/x²)]e^(−x)/r − (g²/12m²)σ₁⋅σ₂δ³(r), derived. Check it on the singlet, where σ₂ acts as −σ₁: Ṽ = (g²/4m²)|q|²/(|q|² + m_φ²) = (g²/4m²)[1 − m_φ²/(|q|² + m_φ²)], a constant minus a Yukawa, and the r-space result with σ₁⋅σ₂ = −3 and S₁₂ = 0 is exactly (g²/4m²)[δ³(r) − m_φ²Y].

Which spin states feel which term

σ₁⋅σ₂ = 2S(S + 1) − 3 is −3 on the singlet and +1 on the triplet. S₁₂ = 6(S⋅r̂)² − 2S² vanishes on the singlet, which has no direction to point, and averages to zero over directions of r̂. For both spins along z with r̂ at angle θ, ⟨S₁₂⟩ = 3cos²θ − 1: +2 end to end, −1 side by side, 0 at the magic angle 54.7°. With g²/4π = 13.6, m = 939 MeV and m_φ = 135 MeV the strength is V₀ = (g²/4π)m_φ³/12m² = 3.16 MeV and the range ħ/m_φc = 1.46 fm. At r = 1.5 fm, x = 1.03, V₀e^(−x)/x = 1.10 MeV and T = 6.77, so the singlet sits at −3.31 MeV while |↑↑⟩ gives +16.1 MeV end to end and −6.37 MeV side by side. End to end the tensor part is 2T = 13.5 times the spin-spin part: the force on a polarised pair is mostly tensor, and one pair at one distance is pushed or pulled by its orientation alone. Its zero falls at 57.8°, just past the magic angle, because the spin-spin +1 needs ⟨S₁₂⟩ = −1/T to cancel it.

Checks: dimensions, the massless limit and magnets

Dimensions: with [ψ] = 3/2 and [φ] = 1, g is dimensionless, so (g²/4π)(m_φ²/m²)e^(−x)/r has mass dimension 1, an energy, and so does (g²/m²)δ³(r), with 3 − 2 = 1. Massless limit: as m_φr → 0 the spin-spin Yukawa term vanishes as m_φ² while the tensor term's m_φ²(3/x²)e^(−x)/r tends to 3/r³, leaving V → (g²/4π)S₁₂/(4m²r³) plus the contact term, derived: a pure dipole-dipole law. Two Dirac magnetic moments −μ_Bσ have energy (μ₀/4πr³)[μ₁⋅μ₂ − 3(μ₁⋅r̂)(μ₂⋅r̂)] = −(α/4m²)S₁₂/r³, because μ₀μ_B²/4π = α/4m² in natural units, quoted. Same 1/r³, same angular shape, opposite sign: a spin-0 field attracts like sources, so its dipoles repel end to end, where bar magnets attract. Non-relativistic limit: the result is the leading term in |q|/m, with corrections of relative order |q|²/m² dropped, so it describes separations r ≫ 1/m, 0.21 fm for a nucleon, and the contact term at r = 0 lies outside its reach.

One-pion exchange and the deuteron, with quoted inputs

Real nucleons carry isospin. The coupling −igψ̄γ⁵τ⋅πψ (quoted) multiplies the potential by τ₁⋅τ₂, which is +1 for isospin T = 1 and −3 for T = 0, and the pion-nucleon coupling g²/4π ≈ 13.6 is quoted from πN and NN data; on shell it equals the pseudovector coupling with f/m_π = g/2m, so f²/4π ≈ 0.075, also quoted. The nucleons stay point Dirac fermions. The deuteron has S = 1 and T = 0, so the −3 reverses every sign: the tensor term now attracts end to end and repels side by side, and the deuteron is stretched along its spin. Its measured quadrupole moment, Q = +0.286 fm², positive and so prolate, agrees in sign. Because S₁₂ averages to zero in an S wave, the tensor force acts by mixing: S₁₂ acting on ³S₁ gives √8 times ³D₁ (quoted), the S-D mixing UAQM-02.04 named, a few per cent of the ground state in realistic models. One-pion exchange is the long-range tail, trusted beyond about 2 fm; inside, two-pion, ρ and ω exchange and the nucleons' size take over.

Change one variable at a time

Make the relationship visible.

Interactive model

Angle θ of r̂ from the spins0 °

Boson mass m_φc²135 MeV

Marked separation r₀1.5 fm

Sweep θ from 0° to 90°: the solid curve swings from repulsion to attraction while the dashed singlet never moves. Then drag m_φ to 0: the singlet's attraction vanishes and the solid curve becomes the pure 1/r³ dipole law, with its zero pinned at the magic angle 54.7°.

⟨S₁₂⟩ FOR ↑↑ AT θ2.00

V↑↑ AT r₀16.1 MeV

V SINGLET AT r₀-3.31 MeV

ZERO OF V↑↑ AT r₀: θ₀57.8 °

Live interpretation⟨S₁₂⟩ FOR ↑↑ AT θ: 2.00. V↑↑ AT r₀: 16.1 MeV. V SINGLET AT r₀: −3.31 MeV. ZERO OF V↑↑ AT r₀: θ₀: 57.8 °

Catch the common trap

Explain before calculating.

Misconception check

The tempting argument: the exchanged boson has spin 0, so its exchange must give a central force, a Yukawa like the scalar's with at most a different sign. It fails at the vertex, not the propagator. The boson's spin is zero but its parity is −1, so a slow fermion can emit it only in a P wave, and ū(p′)γ⁵u(p) = χ′†σ⋅(p − p′)χ has no spin-independent piece at all. Spin and direction enter through the vertices, and two of them make (σ₁⋅q)(σ₂⋅q), whose traceless part is the tensor force. Taken seriously, a central Yukawa with g²/4π = 13.6 and m_φ = 135 MeV would be −641 MeV at 1.5 fm, eighty times the binding energy per nucleon. The real force there is −3.3 MeV in the singlet and +16.1 or −6.4 MeV for a polarised pair, set by its orientation alone, and it averages to zero over unpolarised spins.

Two identical point fermions with both spins along +z interact by pseudoscalar exchange, V(r) = V₀(e^(−x)/x)[σ₁⋅σ₂ + S₁₂(1 + 3/x + 3/x²)] with V₀ positive, at a separation where x = m_φr = 1. How does V depend on the direction r̂ of their separation?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA point Dirac fermion of mass 0.938 GeV, spin up along z, moves along +z with |p| = 0.150 GeV in the centre-of-mass frame and scatters elastically through 90° into the +x direction. Evaluate ū(p′)γ⁵u(p) for a final spin up and for a final spin down along z, and compare each with ū(p′)u(p) for the same final spin.

Elastic in the centre of mass, so E′ = E and ū(p′)γ⁵u(p) = χ′†σ⋅(p − p′)χ exactly. Here a = p − p′ = (−0.150, 0, 0.150) GeV, so |q| = 0.150√2 GeV = 0.212 GeV.
No flip: ⟨↑|σ⋅a|↑⟩ = a_z, so ū(p′)γ⁵u(p) = +0.150 GeV. Flip: ⟨↓|σ⋅a|↑⟩ = aₓ + ia_y, so ū(p′)γ⁵u(p) = −0.150 GeV. Check: 0.150² + 0.150² = 0.0450 GeV² = |q|², the whole density shared between the two final spins.
Scalar, from the same spinors (UAQM-04, quoted): ū(p′)u(p) = (E + m)χ′†χ − χ′†[p′⋅p + iσ⋅(p′ × p)]χ/(E + m). Here E = √(0.938² + 0.150²) GeV = 0.9499 GeV, p′⋅p = 0 and p′ × p = −0.0225 GeV² along y.
No flip: ⟨↑|σ_y|↑⟩ = 0, so ū(p′)u(p) = E + m = 1.888 GeV, 0.6% above 2m = 1.876 GeV. Flip: ⟨↓|σ_y|↑⟩ = i, so ū(p′)u(p) = −i(−0.0225)(i)/1.888 GeV = −0.0119 GeV.
The scalar density is almost all 2m and almost all no-flip; the pseudoscalar density is all spin, of size |q|, and 0.150/1.888 = 8% of the scalar's here.

Answerū(p′)γ⁵u(p) = +0.150 GeV without spin flip and −0.150 GeV with it, of combined size |q| = 0.212 GeV; ū(p′)u(p) = 1.888 GeV and −0.0119 GeV.

MediumIn the deuteron's channel a neutron and a proton have S = 1 and T = 0, so one-pion exchange is the point-fermion potential times τ₁⋅τ₂ = −3 (quoted). Take g²/4π = 13.6, mc² = 938.9 MeV and m_πc² = 138.0 MeV (isospin averages, quoted). With both spins along +z at r = 2.00 fm, find V with r̂ along z and with r̂ perpendicular to z, then its average over directions of r̂, and say which deuteron shape the force favours.

Strength: V₀ = (g²/4π)(m_πc²)³/(12(mc²)²) = 13.6 × 138.0³/(12 × 938.9²) MeV = 3.379 MeV, with range ħc/m_πc² = 1.430 fm.
Radial factors: x = 138.0 × 2.00/197.327 = 1.399, so V₀e^(−x)/x = 3.379 × 0.1765 MeV = 0.5965 MeV, and T = 1 + 3/x + 3/x² = 1 + 2.145 + 1.533 = 4.678.
Along z, ⟨σ₁⋅σ₂⟩ = 1 and ⟨S₁₂⟩ = 2: V = −3 × 0.5965 × (1 + 2 × 4.678) MeV = −18.5 MeV. Perpendicular, ⟨S₁₂⟩ = −1: V = −3 × 0.5965 × (1 − 4.678) MeV = +6.58 MeV.
Over directions ⟨S₁₂⟩ averages to 0, leaving −3 × 0.5965 MeV = −1.79 MeV. That is also the np ¹S₀ value, where σ₁⋅σ₂ = −3 and τ₁⋅τ₂ = +1: in a pure S wave the tensor force does nothing at first order.
The energy is lowest with the pair lined up along its spin, so the tensor force mixes in the D wave that stretches the deuteron along its spin axis: prolate, Q positive, as the measured Q = +0.286 fm² confirms in sign.

AnswerV = −18.5 MeV along the spin axis and +6.58 MeV across it, averaging to −1.79 MeV over directions; the force favours a prolate deuteron, as the measured Q = +0.286 fm² confirms.

HardA hypothetical pseudoscalar of mass m_φc² = 1.00 μeV couples to electrons through igφψ̄γ⁵ψ with g = 1.00 × 10⁻¹³. Two electrons with both spins along +z sit on the z axis. Show that for m_φr ≪ 1 their pseudoscalar potential has the shape of their magnetic dipole-dipole energy, find the ratio of the two at r = 30.0 cm, and find the separation at which the boson's mass has halved that ratio.

On the axis ⟨σ₁⋅σ₂⟩ = 1 and ⟨S₁₂⟩ = 2, so the derived potential becomes V = (g²/4π)(m_φ²/12m²)[1 + 2(1 + 3/x + 3/x²)]e^(−x)/r = (g²/4π)e^(−x)(2 + 2x + x²)/(4m²r³), with x = m_φr.
Magnetic, quoted: Dirac moments −μ_Bσ give U = (μ₀/4πr³)[μ₁⋅μ₂ − 3(μ₁⋅ẑ)(μ₂⋅ẑ)] = −2μ₀μ_B²/(4πr³), and μ₀μ_B²/4π = αħ³/(4m²c) = 8.60 × 10⁻⁵⁴ J m³, so in natural units U = −2α/(4m²r³).
Ratio: V/U = −[(g²/4π)/α] e^(−x)(1 + x + x²/2). As x → 0 the last factor tends to 1 at every r: the same 1/r³ and the same S₁₂ shape, opposite sign. With g²/4π = 7.96 × 10⁻²⁸ the massless ratio is −1.09 × 10⁻²⁵.
Range: ħc/m_φc² = 197.327 MeV fm ÷ (1.00 × 10⁻¹² MeV) = 0.1973 m. At r = 0.300 m, x = 1.520 and e^(−x)(1 + x + x²/2) = 0.2186 × 3.676 = 0.804, so V/U = −8.76 × 10⁻²⁶.
Halving: solve e^(−x)(1 + x + x²/2) = 1/2 by bisection, giving x = 2.674 and r = 2.674 × 0.1973 m = 0.528 m.

AnswerV/U = −1.09 × 10⁻²⁵ × e^(−x)(1 + x + x²/2): −8.76 × 10⁻²⁶ at 30.0 cm, repulsive where the magnets attract. The mass halves the ratio at r = 0.528 m.